Lateral Self Equilibrium Example

Consider the structure with two elevated floors shown in the following figure. Each level is 3m high and the structure is 10m wide.

Example with two elevated floors

Assume the following:

a frame analysis has been performed on the building for this 100kN loading and the column forces are known
a very simple distribution of forces (reasonable for beams much stiffer than columns)

The forces on the top level slab (including column reactions) are:

Forces on top level slab

F_x0 = 100kN
F_x1 = -50kN	F_x2 = -50kN
F_z1 = -15kN	F_z2 = 15kN
M_y1 = 75kN-m	M_y2 = 75kN-m

These forces are in equilibrium and are applied directly to the slab in a lateral SE loading. RAM Concept then calculates the correct forces in the slab, design strips and punching checks.

For the intermediate level there are more forces to consider (all of these are from the frame analysis). The forces that the columns apply to the slab are:

Forces on intermediate level slab

F_x3 = 50kN	F_x4 = -50kN
F_x5 = 50kN	F_x6 = -50kN
F_z3 = 15kN	F_z4 = -45kN
F_z5 = -15kN	F_z6 = 45kN
M_y3 = 75kN-m	M_y4 = 75kN-m
M_y5 = 75kN-m	M_y6 = 75kN-m

These forces are in equilibrium and are applied directly to the slab in a lateral SE loading.

Since the "3" and "4" forces occur at the same location, they can be added together and applied as a single load (same for "5" and "6").

RAM Concept then calculates the correct forces in the slab, design strips and punching checks.

Note: There is one simplification - if you do not care about diaphragm forces, then you can ignore all the Fx and Fy forces. This assumes that the Fx and Fy forces act at the center of your slab and that the centroid elevation of your slab is constant. When these two assumptions are not true, the effects of these forces are typically still not large, but you may need to use some judgment before you ignore them.